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What is trigonometry?
Angle measurement and tables
If there is anything that distinguishes
trigonometry from the rest of geometry,
it is that trig depends on angle measurement
and quantities determined by the measure of an
angle.
Applications of trigonometry
What can you do with trig?
 Historically, it was developed for astronomy and geography, but
scientists have been using it for centuries for other purposes,
too.
 Besides other fields of mathematics, trig is used in physics,
engineering, and chemistry.
 Within mathematics, trig is used in primarily in calculus (which is
perhaps its greatest application), linear algebra, and statistics.
 Since these fields are used throughout the natural and social
sciences, trig is a very useful subject to know
From the information given in the diagram for each
question, find:
tan A = 1.2424242
A = 51.170175
1. Angle
A to the
A = 510degree
nearest
Q
A
P
8.9cm
R
33cm
B
13.7cm
sin P =P 0.649635
2. Angle
to 1dp
C
P = 40.514091
41 cm
Cos 35 = z÷23.6
P = 40.50 (1dp)
z = 23.6 x 0.819152
3. The length of
z = 19.331988
Y
side z to 3 sf
z = 19.3cm(3sf)
Tan 22 = 14 ÷ b
z
b
A
220
14cm
Tan 22 x b = 14
350
X
C
23.6 cm
b = 14 ÷ 0.4040262
B
Z
b = 34.651216
4. The length of b to 2 sf.
b = 35cm (2sf)
Angles of elevation and depression
Many practical problems involve the use of right-angled
triangles.
Simple versions involve:
 Angle of elevation
 Angle of depression
Example
Find the height of a tree if someone standing 12 m from the base
measures the angle of elevation of the top of the tree to be 350 .
h
350
H
12 m
tan H = opposite ÷ adjacent
tan 35 = h ÷ 12
h = 12 tan 35
h = 8.4024905
h = 8.4 m (2sf)
The height of the tree is 8.4 m
…draw a diagram
…decide what sides are involved
…write appropriate formula
…substitute
…make h the subject
Exercise
1.
An 8.0 m ladder is leaning against a wall, with the base of the
ladder 2.0 m from the wall. What is the angle of elevation of
the top of the ladder from the base?
The ladder makes an angle of 75.50 with the ground.
2.
Peak B is 85 m above peak A. Someone is standing at peak B
measured the angle of depression of peak A to be 280 . What
is the direct distance between A and B?
The distance between A and B is 180m
3.
A yachtie 500 m from the base of a cliff measures the angle of
elevation of the top of the cliff to be 170
a) How high is the cliff?
The height of the cliff is 153 m
b) What is the new angle of elevation if the yacht moves away
another 150m
The angle of elevation is 130
Bearings
The bearing of A from B is the angle
between the north line through
B, and the line AB, measured
clockwise from the north line.
So A has a bearing of 0550 from B
N
A
550
B
N
In the next diagram,
a) H has a bearing of 1020 from E
b) E has a bearing of 2820 from H
E
N
1020
H
2820
N
0750 on
A woman walks 330 m on a bearing of
horizontal ground
a) How far north of her starting point has she travelled?
b) What bearing would she take to return to her starting point?
B
a) Draw a diagram using a right angled triangle.
a is the adjacent side to the 750 angle so use
cos 75 = a
a
330
a = 330 x cos 75
a = 85.410285
a = 85m (2sf)
The woman is 85m north of her starting point
b) Redraw the diagram, with a north line
N
at her “end” point.
By using geometry,
x = 1050 …co-interior angles, parallal lines
y = 2550 …angles at a point
Her bearing back to start is 2550.
750
750
330m
330m
N
750
x y
Exercise
1. An aeroplane travels at a distance of 120 km
on a course of 0600. How far east of its
starting point is it?
2. A ship travels south for 2.7 km, then east for
3.5 km.
a) How far is it from its starting point?
b) What is the bearing of the final position
from the starting point?
An aeroplane travels at a distance of 120 km on a course of 0600.
How far east of its starting point is it?
 Given hypotenuse = 120 km angle A = 600
to find o (the distance east)
N
Use sin A = o
h
sin 60 = o
600
120
o = 120 x sin 60
A
o = 103.92305
o = 104 km (3sf)
The plane is 104 km east of its starting point
o
B
120
A ship travels south for 2.7 km, then east for 3.5 km.
a) How far is it from its starting point?
b) What is the bearing of the final position from the starting point?
a)
Given two sides, to find the third side, use Pythagoras
A
b2
b2
b
b
= 2.72 + 3.52
= 19.54
= 4.4204072
= 4.4 km (1dp)
2.7
B
The ship is 4.4 km from the starting point
b
3.5
C
A ship travels south for 2.7 km, then east for 3.5 km.
a) How far is it from its starting point?
b) What is the bearing of the final position from the starting point?
b) The bearing of C from A is shown. Need to find angle BAC
first.
Given adjacent side = 2.7 km
N
Opposite side = 3.5 km.
A
to find angle A use
tan A = 3.5
2 .7
2.7
b
tan A = 1.2962963
A = 52.352379
bearing = 180 – A
= 127.60 (1dp)
The bearing of C from A is 127.60
B
3.5
C
Non right-angled triangles
Sine and cosine of obtuse angles (900 – 1800)
sin 1250 = 0.819152
cos 1250 = -0.5735764
To find θ given cos θ
If cos θ is negative the angle will be obtuse
cos θ = – 0.602
θ = 127.01327
θ = 1270 (nearest degree)
To find θ given sin θ
Sine is positive for both acute and obtuse angles
sin θ = 0.257
θ = 14.892127 (acute angle given by calculator)
180 - 14.892127 = 165.10787 (obtuse angle)
θ = 150 or 1650 (nearest degree)
Remember
For an angle θ in a triangle,
 If 00 < θ < 1800 and cos θ is positive, the angle θ is acute
 If 00 < θ < 1800 and cos θ is negative, the angle θ is obtuse
 If 00 < θ < 1800 and sin θ is always positive, so angle θ may be
acute or obtuse
The cosine rule
The cosine rule may be used to find:
- the third side of a triangle when you are given the two sides
and the included angle (SAS)
- any angle of a triangle when you are given three sides (SSS)
The cosine rule to find a side
If b and c and angle A are known then
a2 = b2 + c2 – 2bc cos A
Notice that the side of length a is opposite the angle A, and
similarly for b and c
The cosine rule
a2 = b2 + c2 – 2bc cos A
unknown side
known lengths
angle opposite
unknown side
(included angle)
a
Find the length of a.
The Answer
b = 7cm
…draw the diagram
c = 3 cm
A = 350
…list the information
a2 = b2 + c2 – 2bc cos A
…write the formula
 a2 = 32 + 72 - 2 x 3 x 7 x cos 35°
…substitute and evaluate
= 9 + 49 - 42 cos 35°
= 58 - 42 cos 35°
= 58 - 34.404
= 23.5956...
a = √23.5956
= 4.857...
= 4.86 cm (2dp)
Angle A in the triangle is 45o, length b is 2 units and length c
is 3 units. Find all the unknown lengths and angles.
The Answer
b = 2cm
…draw the diagram
c = 3 cm
A = 450
3
To work out the remaining length
a2 = b2 + c2 – 2bc cos A
a2 = 4 + 9 -12 cos(45o), (≈ 2.1247867248)
a is 2.1units (2dp)
A
B
450
2
a
C
To work out the angle B, rewrite the cosine rule so that the angle is B:
b2 = a2 + c2 - 2ac cos B
Now we can substitute in our values for a, b and c, to find cos B:
4 = 4.5 + 9 - 12.6 cos B
cos B = 0.75
( cos-1 (0.7463273462) ≈ 41.72676505)
B is 41.1o (1dp)
Finally we can work out the angle C by remembering that the three angles
must add up to 180o.
This gives C to be 93.9o (1dp)
Find the size of angle R.
The Answer
r = 4 cm
q = 6.9 cm
p = 4.2 cm
…list the information
r2 = q2 + p2 – 2qp cos R
…write the formula
42 = 6.92 + 4.22 - 2 x 6.9 x 4.2 x cos R°
…substitute and evaluate
42 – 6.92 – 4.22 = - 2 x 6.9 x 4.2 x cos R
42 – 6.92 – 4.22 = cos R
- 2 x 6.9 x 4.2
Taking inverse cosine: R = 31.8° (1dp)
Mixed problems
Question 1
 A surveyor wants to find the height of a hill.
 Going back from its base on level ground, he
measures the angle of elevation to the top of the hill
as 31°.
 Moving back a further 130m, he measures the
angle of elevation to be 23°.
 Find the height of the hill.
Question 2
•Town B is 23km north of town A and town C is 29km
south east of town A
•Find the distance between towns B and C